Solve Problem: Arrow Shot Horizontally Hitting Target 25m Away

[mystry4]

I need to make sure I am using the correct formulas (dvertical = 1/2at^2 solving for t^2 and dhorizontal = Vot) for the following problem:
arrow shot horizontally - speed of 100m/s - at bullseye 25 m away. How long til it reaches target and where will it hit target.
Thanks!

 

[HallsofIvy]

Yes, if the arrow is shot horizontally, those are the correct equations. Set dhorizontal= 25 m, a= -9.81 m/s², and Vo= 100 m/s. Solve the dhorizontal equation for t (how long till it reaches the target) and then use that value to solve for dvertical (where will it hit the target).

 

[thepatientmental]

To solve this problem, we can use the equations of motion to determine the time it takes for the arrow to reach the target and where it will hit the target.

First, let's define our variables:

- dvertical: vertical displacement (in this case, it is 0 since the arrow is shot horizontally)
- dhorizontal: horizontal displacement (25 m)
- a: acceleration due to gravity (9.8 m/s^2)
- t: time (what we are trying to find)
- Vo: initial velocity (100 m/s)

Using the equation dhorizontal = Vot, we can rearrange it to solve for t:

t = dhorizontal / Vo

Substituting our values, we get:

t = 25 m / 100 m/s = 0.25 s

So it will take 0.25 seconds for the arrow to reach the target.

To find where it will hit the target, we can use the equation dvertical = 1/2at^2 and solve for dvertical:

dvertical = 1/2 * 9.8 m/s^2 * (0.25 s)^2 = 0.30625 m

Therefore, the arrow will hit the target 0.30625 meters below the bullseye.

Just a note, we assumed that the arrow was shot at ground level and there is no air resistance. If we take into account the height at which the arrow is shot and air resistance, the calculations may be slightly different. But for this problem, these equations should give a reasonable estimate.